A photodetector is made from a compound semiconductor with band gap 0.73eV. The maximum wavelength it can detect is

Options

A. 12400 Å
B. 17030 Å
C. 6200 Å
D. 1703 Å

Answer

To find the maximum wavelength (\( \lambda_{max} \)) that a photodetector can detect, we can follow these steps:

### Step 1: Understand the relationship between energy and wavelength
The energy (\( E \)) of a photon is related to its wavelength (\( \lambda \)) by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)).

### Step 2: Identify the minimum energy required
The minimum energy that the photodetector can detect is given by the band gap energy, which is \( 0.73 \, \text{eV} \). To use this in our equation, we need to convert this energy into joules:
\[ E = 0.73 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.168 \times 10^{-19} \, \text{J} \]

### Step 3: Rearrange the energy-wavelength equation
To find the maximum wavelength, we rearrange the equation for wavelength:
\[ \lambda = \frac{hc}{E} \]

### Step 4: Substitute the values
Now we can substitute the values of \( h \), \( c \), and \( E \):
\[ \lambda_{max} = \frac{(6.63 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{1.168 \times 10^{-19} \, \text{J}} \]

### Step 5: Calculate \( \lambda_{max} \)
Calculating the above expression:
\[ \lambda_{max} = \frac{1.989 \times 10^{-25}}{1.168 \times 10^{-19}} \approx 1.70 \times 10^{-6} \, \text{m} \]
To convert this to nanometers:
\[ \lambda_{max} \approx 17030 \, \text{nm} \text{ (or } 17030 \times 10^{-10} \text{ m)} \]

### Final Answer
Thus, the maximum wavelength that the photodetector can detect is approximately:
\[ \lambda_{max} \approx 17030 \, \text{nm} \]